Aljabar Linear Contoh

2 x - 3 y + 4 z = 4

$2 x - 3 y + 4 z = 4$ ,

- 4 x + y - 3 z = 3

$- 4 x + y - 3 z = 3$ ,

2 x + 2 y - z = 1

$2 x + 2 y - z = 1$

Step 1

Tentukan

A X = B

$A X = B$ dari sistem persamaan tersebut.

[\begin{array}{c} 2 & - 3 & 4 \\ - 4 & 1 & - 3 \\ 2 & 2 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 3 \\ 1 \end{array}]

$[\begin{array}{c} 2 & - 3 & 4 \\ - 4 & 1 & - 3 \\ 2 & 2 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 3 \\ 1 \end{array}]$

Step 2

Tentukan balikan dari matriks koefisien.

Ketuk untuk lebih banyak langkah...

Buat sebuah matriks yang dibagi menjadi dua bagian berukuran sama. Di sisi kiri, masukkan elemen matriks asli. Di sisi kanan, masukkan elemen matriks satuan. Untuk menemukan matriks balikan, gunakan operasi baris untuk mengonversikan sisi kiri ke dalam matriks satuan. Setelah selesai, matriks balikan asli akan berada di sisi kanan matriks ganda.

[\begin{array}{c} 2 & - 3 & 4 & 1 & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 2 & - 3 & 4 & 1 & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{1} = \frac{1}{2} R_{1}

$R_{1} = \frac{1}{2} R_{1}$ pada

R_{1}

$R_{1}$ (baris

1

$1$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

1

$1$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan operasi baris

R_{1} = \frac{1}{2} R_{1}

$R_{1} = \frac{1}{2} R_{1}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

1

$1$ .

[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

R_{1} = \frac{1}{2} R_{1}

$R_{1} = \frac{1}{2} R_{1}$

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{1} = \frac{1}{2} R_{1}

$R_{1} = \frac{1}{2} R_{1}$ .

[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (- 3) & (\frac{1}{2}) \cdot (4) & (\frac{1}{2}) \cdot (1) & (\frac{1}{2}) \cdot (0) & (\frac{1}{2}) \cdot (0) \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (- 3) & (\frac{1}{2}) \cdot (4) & (\frac{1}{2}) \cdot (1) & (\frac{1}{2}) \cdot (0) & (\frac{1}{2}) \cdot (0) \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

R_{1} = \frac{1}{2} R_{1}

$R_{1} = \frac{1}{2} R_{1}$

Sederhanakan

R_{1}

$R_{1}$ (baris

1

$1$ ).

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - 4 & 1 & - 3 & 0 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{2} = 4 \cdot R_{1} + R_{2}

$R_{2} = 4 \cdot R_{1} + R_{2}$ pada

R_{2}

$R_{2}$ (baris

2

$2$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan operasi baris

R_{2} = 4 \cdot R_{1} + R_{2}

$R_{2} = 4 \cdot R_{1} + R_{2}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

R_{2} = 4 \cdot R_{1} + R_{2}

$R_{2} = 4 \cdot R_{1} + R_{2}$

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{2} = 4 \cdot R_{1} + R_{2}

$R_{2} = 4 \cdot R_{1} + R_{2}$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ (4) \cdot (1) - 4 & (4) \cdot (- \frac{3}{2}) + 1 & (4) \cdot (2) - 3 & (4) \cdot (\frac{1}{2}) + 0 & (4) \cdot (0) + 1 & (4) \cdot (0) + 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ (4) \cdot (1) - 4 & (4) \cdot (- \frac{3}{2}) + 1 & (4) \cdot (2) - 3 & (4) \cdot (\frac{1}{2}) + 0 & (4) \cdot (0) + 1 & (4) \cdot (0) + 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

R_{2} = 4 \cdot R_{1} + R_{2}

$R_{2} = 4 \cdot R_{1} + R_{2}$

Sederhanakan

R_{2}

$R_{2}$ (baris

2

$2$ ).

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 2 & 2 & - 1 & 0 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{3} = - 2 \cdot R_{1} + R_{3}

$R_{3} = - 2 \cdot R_{1} + R_{3}$ pada

R_{3}

$R_{3}$ (baris

3

$3$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{3}

$R_{3}$ (baris

3

$3$ ) dengan operasi baris

R_{3} = - 2 \cdot R_{1} + R_{3}

$R_{3} = - 2 \cdot R_{1} + R_{3}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} \end{array}]$

R_{3} = - 2 \cdot R_{1} + R_{3}

$R_{3} = - 2 \cdot R_{1} + R_{3}$

Ganti

R_{3}

$R_{3}$ (baris

3

$3$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{3} = - 2 \cdot R_{1} + R_{3}

$R_{3} = - 2 \cdot R_{1} + R_{3}$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- \frac{3}{2}) + 2 & (- 2) \cdot (2) - 1 & (- 2) \cdot (\frac{1}{2}) + 0 & (- 2) \cdot (0) + 0 & (- 2) \cdot (0) + 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- \frac{3}{2}) + 2 & (- 2) \cdot (2) - 1 & (- 2) \cdot (\frac{1}{2}) + 0 & (- 2) \cdot (0) + 0 & (- 2) \cdot (0) + 1 \end{array}]$

R_{3} = - 2 \cdot R_{1} + R_{3}

$R_{3} = - 2 \cdot R_{1} + R_{3}$

Sederhanakan

R_{3}

$R_{3}$ (baris

3

$3$ ).

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & - 5 & 5 & 2 & 1 & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ pada

R_{2}

$R_{2}$ (baris

2

$2$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

1

$1$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan operasi baris

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

1

$1$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$ .

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (5) & (- \frac{1}{5}) \cdot (2) & (- \frac{1}{5}) \cdot (1) & (- \frac{1}{5}) \cdot (0) \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (5) & (- \frac{1}{5}) \cdot (2) & (- \frac{1}{5}) \cdot (1) & (- \frac{1}{5}) \cdot (0) \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

R_{2} = - \frac{1}{5} R_{2}

$R_{2} = - \frac{1}{5} R_{2}$

Sederhanakan

R_{2}

$R_{2}$ (baris

2

$2$ ).

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{3}{2} & 2 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{1} = \frac{3}{2} R_{2} + R_{1}

$R_{1} = \frac{3}{2} R_{2} + R_{1}$ pada

R_{1}

$R_{1}$ (baris

1

$1$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan operasi baris

R_{1} = \frac{3}{2} R_{2} + R_{1}

$R_{1} = \frac{3}{2} R_{2} + R_{1}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} & \frac{3}{2} R_{2} + R_{1} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

R_{1} = \frac{3}{2} R_{2} + R_{1}

$R_{1} = \frac{3}{2} R_{2} + R_{1}$

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{1} = \frac{3}{2} R_{2} + R_{1}

$R_{1} = \frac{3}{2} R_{2} + R_{1}$ .

[\begin{array}{c} (\frac{3}{2}) \cdot (0) + 1 & (\frac{3}{2}) \cdot (1) - \frac{3}{2} & (\frac{3}{2}) \cdot (- 1) + 2 & (\frac{3}{2}) \cdot (- \frac{2}{5}) + \frac{1}{2} & (\frac{3}{2}) \cdot (- \frac{1}{5}) + 0 & (\frac{3}{2}) \cdot (0) + 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} (\frac{3}{2}) \cdot (0) + 1 & (\frac{3}{2}) \cdot (1) - \frac{3}{2} & (\frac{3}{2}) \cdot (- 1) + 2 & (\frac{3}{2}) \cdot (- \frac{2}{5}) + \frac{1}{2} & (\frac{3}{2}) \cdot (- \frac{1}{5}) + 0 & (\frac{3}{2}) \cdot (0) + 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

R_{1} = \frac{3}{2} R_{2} + R_{1}

$R_{1} = \frac{3}{2} R_{2} + R_{1}$

Sederhanakan

R_{1}

$R_{1}$ (baris

1

$1$ ).

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 5 & - 5 & - 1 & 0 & 1 \end{array}]$

Lakukan operasi baris

R_{3} = - 5 \cdot R_{2} + R_{3}

$R_{3} = - 5 \cdot R_{2} + R_{3}$ pada

R_{3}

$R_{3}$ (baris

3

$3$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{3}

$R_{3}$ (baris

3

$3$ ) dengan operasi baris

R_{3} = - 5 \cdot R_{2} + R_{3}

$R_{3} = - 5 \cdot R_{2} + R_{3}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} \end{array}]$

R_{3} = - 5 \cdot R_{2} + R_{3}

$R_{3} = - 5 \cdot R_{2} + R_{3}$

Ganti

R_{3}

$R_{3}$ (baris

3

$3$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{3} = - 5 \cdot R_{2} + R_{3}

$R_{3} = - 5 \cdot R_{2} + R_{3}$ .

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ (- 5) \cdot (0) + 0 & (- 5) \cdot (1) + 5 & (- 5) \cdot (- 1) - 5 & (- 5) \cdot (- \frac{2}{5}) - 1 & (- 5) \cdot (- \frac{1}{5}) + 0 & (- 5) \cdot (0) + 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ (- 5) \cdot (0) + 0 & (- 5) \cdot (1) + 5 & (- 5) \cdot (- 1) - 5 & (- 5) \cdot (- \frac{2}{5}) - 1 & (- 5) \cdot (- \frac{1}{5}) + 0 & (- 5) \cdot (0) + 1 \end{array}]$

R_{3} = - 5 \cdot R_{2} + R_{3}

$R_{3} = - 5 \cdot R_{2} + R_{3}$

Sederhanakan

R_{3}

$R_{3}$ (baris

3

$3$ ).

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{10} & - \frac{3}{10} & 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

Lakukan operasi baris

R_{1} = \frac{1}{10} R_{3} + R_{1}

$R_{1} = \frac{1}{10} R_{3} + R_{1}$ pada

R_{1}

$R_{1}$ (baris

1

$1$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan operasi baris

R_{1} = \frac{1}{10} R_{3} + R_{1}

$R_{1} = \frac{1}{10} R_{3} + R_{1}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} & \frac{1}{10} R_{3} + R_{1} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

R_{1} = \frac{1}{10} R_{3} + R_{1}

$R_{1} = \frac{1}{10} R_{3} + R_{1}$

Ganti

R_{1}

$R_{1}$ (baris

1

$1$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{1} = \frac{1}{10} R_{3} + R_{1}

$R_{1} = \frac{1}{10} R_{3} + R_{1}$ .

[\begin{array}{c} (\frac{1}{10}) \cdot (0) + 1 & (\frac{1}{10}) \cdot (0) + 0 & (\frac{1}{10}) \cdot (0) + \frac{1}{2} & (\frac{1}{10}) \cdot (1) - \frac{1}{10} & (\frac{1}{10}) \cdot (1) - \frac{3}{10} & (\frac{1}{10}) \cdot (1) + 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} (\frac{1}{10}) \cdot (0) + 1 & (\frac{1}{10}) \cdot (0) + 0 & (\frac{1}{10}) \cdot (0) + \frac{1}{2} & (\frac{1}{10}) \cdot (1) - \frac{1}{10} & (\frac{1}{10}) \cdot (1) - \frac{3}{10} & (\frac{1}{10}) \cdot (1) + 0 \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

R_{1} = \frac{1}{10} R_{3} + R_{1}

$R_{1} = \frac{1}{10} R_{3} + R_{1}$

Sederhanakan

R_{1}

$R_{1}$ (baris

1

$1$ ).

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & - \frac{2}{5} & - \frac{1}{5} & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

Lakukan operasi baris

R_{2} = \frac{2}{5} R_{3} + R_{2}

$R_{2} = \frac{2}{5} R_{3} + R_{2}$ pada

R_{2}

$R_{2}$ (baris

2

$2$ ) untuk mengubah beberapa elemen dalam baris tersebut menjadi

0

$0$ .

Ketuk untuk lebih banyak langkah...

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan operasi baris

R_{2} = \frac{2}{5} R_{3} + R_{2}

$R_{2} = \frac{2}{5} R_{3} + R_{2}$ dengan tujuan mengubah beberapa elemen dalam baris menjadi nilai yang diinginkan

0

$0$ .

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} & \frac{2}{5} R_{3} + R_{2} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

R_{2} = \frac{2}{5} R_{3} + R_{2}

$R_{2} = \frac{2}{5} R_{3} + R_{2}$

Ganti

R_{2}

$R_{2}$ (baris

2

$2$ ) dengan nilai aktual dari elemen-elemen untuk operasi baris

R_{2} = \frac{2}{5} R_{3} + R_{2}

$R_{2} = \frac{2}{5} R_{3} + R_{2}$ .

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ (\frac{2}{5}) \cdot (0) + 0 & (\frac{2}{5}) \cdot (0) + 1 & (\frac{2}{5}) \cdot (0) - 1 & (\frac{2}{5}) \cdot (1) - \frac{2}{5} & (\frac{2}{5}) \cdot (1) - \frac{1}{5} & (\frac{2}{5}) \cdot (1) + 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ (\frac{2}{5}) \cdot (0) + 0 & (\frac{2}{5}) \cdot (0) + 1 & (\frac{2}{5}) \cdot (0) - 1 & (\frac{2}{5}) \cdot (1) - \frac{2}{5} & (\frac{2}{5}) \cdot (1) - \frac{1}{5} & (\frac{2}{5}) \cdot (1) + 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

R_{2} = \frac{2}{5} R_{3} + R_{2}

$R_{2} = \frac{2}{5} R_{3} + R_{2}$

Sederhanakan

R_{2}

$R_{2}$ (baris

2

$2$ ).

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & - \frac{1}{5} & \frac{1}{10} \\ 0 & 1 & - 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 0 & 0 & 1 & 1 & 1 \end{array}]$

Karena determinan dari matriks tersebut adalah nol, maka tidak ada balikan.

Tidak ada balikan

Step 3

Karena matriks tidak mempunyai balikan, maka matriks tersebut tidak dapat diselesaikan menggunakan matriks balikan.

Tidak ada penyelesaian